Consider a random sample of size $n$ froma gamma distribution, $X_i\sim GAM(\theta, \kappa)$, and let $\bar X=\dfrac{1}{n}\sum X_i$ and $\tilde X=(\prod X_i)^{1/n}$ be the sample mean and geometric mean, respectively.
Show that the distribution of $T=\bar X/\tilde X$ does not depend on $\theta$.
It can be shown easily that $\bar X$ and $\tilde X$ are jointly complete and sufficient statistics for $\theta$ and $\kappa$. I'm not necessarily sure if that will help me with this, but I don't really know how to go about this. Surely I don't need to use bivariate transformation, do I? There's got to be an easier way to show this than that. Any ideas?
$\def\deq{\stackrel{\mathrm{d}}{=}}$Suppose $Y_1, \cdots, Y_n$ are i.i.d. such that $Y_k \sim {\mit Γ}(1, κ)$, then$$ (X_1, \cdots, X_n) \deq (θY_1, \cdots, θY_n). $$ So$$ (\overline{X}, \widetilde{X}) = \left( \frac{1}{n} \sum_{k = 1}^n X_k, \left( \prod_{k = 1}^n X_k \right)^{\frac{1}{n}} \right) \deq \left( θ \cdot \frac{1}{n} \sum_{k = 1}^n Y_k, θ \cdot \left( \prod_{k = 1}^n Y_k \right)^{\frac{1}{n}} \right) = (θ \overline{Y},θ \widetilde{Y}),$$ where$$ \overline{Y} = \frac{1}{n} \sum_{k = 1}^n Y_k,\ \widetilde{Y} = \left( \prod_{k = 1}^n Y_k \right)^{\frac{1}{n}}. $$ Thus$$ \frac{\overline{X}}{\widetilde{X}} \deq \frac{\overline{Y}}{\widetilde{Y}}, $$ which does not depend on $θ$.